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Quantity of an Element in Mixture by Adding another Element

  1. Jun 30, 2017 #1
    1. The problem statement, all variables and given/known data
    A mixture of 12 ounces of vinegar and oil is 40% vinegar, where all of the measurements are by weight. How many ounces of oil must be added to the mixture to produce a new mixture that is only 25 percent vinegar?
    I don’t know how to solve it.

    2. Relevant equations

    May be unitary method

    3. The attempt at a solution
    Let Quantity of oil added = x
    12 ounces = 40% vinegar

    1 ounce = 0.4/12---(1)

    12 + x ounce = 25% vinegar

    1 ounce =. 25/(12 + x)---(2)

    0.4/12 = .25/(12 + x) (Equating 1 & 2)

    But book has different solution.

    Some body please guide me.

    Zulfi.
     
    Last edited by a moderator: Jul 1, 2017
  2. jcsd
  3. Jun 30, 2017 #2

    SammyS

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    What do you mean by the statement:
    " 12 ounces = 40% vinegar " ?​

    What you could do is to determine the amount of vinegar (in ounces) in the initial solution and in the final solution.
     
  4. Jun 30, 2017 #3
    Hi,
    Thanks for your response.
    By the above, i mean that 12 ounces contain 40% vinegar.
    This is what i am doing. I am trying to determine vinegar in one ounce of solution. Maybe my method is not correct.
    Please guide me.

    Zulfi.
     
  5. Jun 30, 2017 #4

    SammyS

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    "12 ounces = 40% vinegar" is meaningless mathematically speaking.

    Clearly, each ounce of the initial mixture has 0.4 ounces of Vinegar. I doubt that this fact is useful.

    So, I repeat:

    How much vinegar ( in ounces ) is in the initial mixture?

    How much vinegar ( in ounces ) is in the final mixture?

    Also: Please respond to your thread https://www.physicsforums.com/threads/selling-price-for-an-item-with-a-markup-and-discount.918917/
     
  6. Jul 1, 2017 #5

    Ray Vickson

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    I will repeat also what SammyS asked you, perhaps in a bit more detail. You have a 12 ounce mixture of oil and vinegar, which is 40% vinegar. How much vinegar and how much oil are present in that bottle of mixture? By "how much" I mean actual numbers of ounces of each.

    Now you add x ounces of oil. How many ounces of mixture do you have now? How many ounces of vinegar do you have now? You ought to be able to take it from there.

    Please note that if you want people to "guide you" you need to pay attention to what they tell you! If you are asked a question, you should answer it. Doing that is how you will learn, because people ask you sensible questions for very good reasons. Many helpers have years of experience teaching the subject and know what hurdles students can face, and know how to help them overcome those hurdles.
     
  7. Jul 1, 2017 #6
    Hi,
    Vinegar is 40% and oil should be 60%. But i dont think that we should discuss about the quantity of oil. In one post in this forum, i was asked to concentrate on thinks mentioned in the question.

    After adding x ounces of oil we have mixture = 12+x ounces
    vinegar = 0.25 of (12 + x) ounces
    My friend i am paying attention. Kindly realize the limitations.
    Yes my friend i know, question-answer method is a very useful method of teaching and it was invented by Aristotle.
    I think we have to equate the initial and final mixtures but how can do that:
    initially vinegar= 0.4 of 12 ounces
    finally vinegar= 0.25 of (12 + x) ounces

    because one is 40% quantity of vinegar and the other is 25% quantity of vinegar.
    I dont think it would be correct to use the quantity of oil for equating the initial and final solutions because they have not stated it.

    Some body please guide me.
    Zulfi.
     
  8. Jul 1, 2017 #7

    SammyS

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    No vinegar was added. No vinegar was removed.

    Why do you have a problem with equating the total amount of vinegar in the initial mixture with the total amount of vinegar in the final mixture?
     
  9. Jul 1, 2017 #8

    Ray Vickson

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    I asked you how much vinegar is present, in actual ounces. What is the answer? Is it 10 ounces or 2 ounces or 4.6 ounces, or what? Are you refusing to answer because you do not really understand percentages and how to use them? If so, just tell us!

    Anyway, the question said you had a mixture of oil and vinegar, so whatever is not vinegar is oil, and that means that you CAN get the amount of oil as well; but as you say, you may not need it.

    No more guidance from me until you deal with my questions.
     
  10. Jul 2, 2017 #9
    Hi,
    Thanks for your interest in my problem. First of all i cant understand what you want me to tell you. Now i would tell you. Secondly there is an unknown x in the final solution. Any way i would try to answer your question.
    Initial = 4.8 ounces
    Final = 5.5 + 0.25x

    In my opinion their quantity is not equal. One is 40% of 12 ounces and other is 25% of (12+x) ounces.

    Please let me know whether i am right or wrong time. Sorry for disturbing you with incorrect answer.

    Zulfi.
     
  11. Jul 3, 2017 #10

    SammyS

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    That is not correct for the final amount.(of vinegar).

    The final amount of vinegar is: 25% of (12+x), Right ?

    Change 25% to its decimal equivalent, and distribute. Then equate the two expressions and solve.
     
  12. Jul 3, 2017 #11

    Ray Vickson

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    Yes, they ARE equal if you choose the correct value of x. That is the entire point of the problem!
     
    Last edited: Jul 3, 2017
  13. Jul 3, 2017 #12
    How many oz of vinegar is present in the original mixture?

    If you add x oz of oil to the mixture, in terms of x, how many oz of mixture do you now have?

    In terms of x, what is the fraction of vinegar present in the new mixture? (This is also equal to 0.25, right?)

    What is the value of x?
     
  14. Jul 5, 2017 #13
    Hi,
    How many oz of vinegar is present in the original mixture? 40% of 12 ounces
    If you add x oz of oil to the mixture, in terms of x, how many oz of mixture do you now have? 12 + x
    In terms of x, what is the fraction of vinegar present in the new mixture? (This is also equal to 0.25, right?): Confusing?? How can you equate 25% to 40%.


    Initial = 4.8 ounces
    Final = 3 + 0.25x ounces
    Maybe this is the reason:
    We are not considering a mixture now. We are considering homogenous solution consisting of only vinegar. So we can equate :
    Okay my friends, we have spent too much time on this.
    I would accept your method.
    God bless you people.

    Zulfi.
     
  15. Jul 5, 2017 #14
    $$\frac{4.8}{12+x}=0.25$$
     
  16. Jul 5, 2017 #15

    Ray Vickson

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    This is far too helpful, and looks like it gives the OP the solution. He has received help over and over and over again, and he ignores it, so I think we should just let him receive his deserved mark on the question (that is, 0).
     
  17. Jul 5, 2017 #16
    You really should be able to translate a sentence like this in mathematical equations, or you'll never be able to do these kinds of problems.
    if o is the amount of oil present initallly and v the amout of vinegar, you can get 2 equations from this sentence
     
  18. Jul 5, 2017 #17
    Thanks Ray. It was a (probably bad) judgment call. In post #13, he had correctly determined the amount of vinegar in the solution (4.8) and he also correctly determined the final total amount of solution (x + 12). He just couldn't make the connection to the final fraction of vinegar (0.25). Anyway, you're right. I probably gave him too much info. Mea culpa.
     
  19. Jul 5, 2017 #18

    scottdave

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    Being a mixture or solution does not change how much vinegar is present. As said by others, the number of ounces of vinegar has not changed (none added or removed). A percentage is a ratio of two quantities.
    Let M0 be the mass of vinegar, and M1 be the initial mass (12 ounces). So a ratio of 40% (or 40/100 = 0.4) is equal to M0 / M1 = M0 / (12 ounces). You can solve for M0 (which you did to be 4.8 ounces).

    The next ratio is 25% (or 0.25). This ratio can be set up with the same amount of vinegar (it didn't change) and the new mass of the total. Then you can solve for the amount which was added.
     
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